On P Versus NP
share
I generalize a well-known result that P = NP fails for monotone polynomial circuits - more precisely, that the clique problem CLIQUE(k^4,k) is not solvable by Boolean (AND,OR)-circuits of the size polynomial in k. In the other words, there is no Boolean (AND,OR)-formula F expressing that a given graph with k^4 vertices contains a clique of k elements, provided that the circuit length of F, cl(F), is polynomial in k. In fact, for any solution F in question, cl(F) must be exponential in k. Moreover this holds also for DeMorgan normal (abbr.: DMN) (AND,OR)-formulas F that allow negated variables. Based on the latter observation I consider an arbitrary (AND,OR,NOT)-formula F and recall that standard NOT-conversions to DMN at most double its circuit length. Hence for any Boolean solution F of CLIQUE(k^4,k), cl(F) is exponential in k. I conclude that CLIQUE(k^4,k) is not solvable by polynomial-size Boolean circuits, and hence P is not NP. The entire proof is formalizable by standard methods in the exponential function arithmetic EFA.
READ FULL TEXT